Mr. McG wants to have more time to teach because he has several students who hav
ID: 3204846 • Letter: M
Question
Mr. McG wants to have more time to teach because he has several students who have mentioned in their concept checks how much they love ECO 391 and how they wish they could have an even longer lecture! However, he is conscious of the fact that some students may have somewhere they have to be following his class. He wants to determine how long it takes students to get to their next activity. He has his buddy Bob Gillette survey 36 of his students. His buddy Bob reports that the confidence interval around the mean time it takes students to get to their next activity is [8.20, 9.80]. Bob is on sabbatical and never provided Mr. McG with the sample mean used to compute this confidence interval or the confidence level. Mr. McG does know that the sample standard deviation is equal to 2.365 – please provide Mr. McG with these missing pieces of information to help him make the right decision!
Explanation / Answer
Result:
Mr. McG wants to have more time to teach because he has several students who have mentioned in their concept checks how much they love ECO 391 and how they wish they could have an even longer lecture! However, he is conscious of the fact that some students may have somewhere they have to be following his class. He wants to determine how long it takes students to get to their next activity. He has his buddy Bob Gillette survey 36 of his students. His buddy Bob reports that the confidence interval around the mean time it takes students to get to their next activity is [8.20, 9.80]. Bob is on sabbatical and never provided Mr. McG with the sample mean used to compute this confidence interval or the confidence level. Mr. McG does know that the sample standard deviation is equal to 2.365 – please provide Mr. McG with these missing pieces of information to help him make the right decision!
n=36
s=2.365
CI=(8.20,9.80)
Mean = ( lower limit+upper limit)/2 = (8.20+9.80)/2 = 9
Margin of error = (upper limit-lower limit)/2 = (9.80-8.20)/2 =0.8
Margin of error = t*sd/sqrt(n)
0.8 =t*2.365/sqrt(36)
t= 0.8*sqrt(36)/2.365 =2.029598
=2.030 ( three decimals)
df =36-1 =35
From t table we found that at 0.05 level of significance with 35 Df is 3.050.
Level of significance = 0.05.
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